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3 years ago

What are the solutions to the equation -54 + 6x = 0

Mathematics

2 answers:

irakobra [83]3 years ago

8 0

Add 54 on both sides of the equation to cancel out the -54 on the left side, the equation 6x = 54

Divide both sides of the equation by 6 to cancel out the 6 in 6x, making the equation x = 9.

There’s the answer. x is equal to 9.

Sladkaya [172]3 years ago

7 0

Answer:

x=9

Step-by-step explanation:

-54 + 6x = 6 • (x - 9)
Solve : 6 = 0
Solve : x-9 = 0
Add 9 to both sides of the equation : x = 9

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Which describes the first step in solving the equation 4x - 7 = -15?

GalinKa [24]

Answer:

Add 7 to both sides

Step-by-step explanation:

You would need to add 7 to both sides to cancel out the 7 on the right side of the equation to get x one step closer to being by itself.

Then you would get 4x = -8.

Next divide by 4 on both sides to get x by itself so do the inverse operation of what 4 was doing to x.

x = -2

7 0

3 years ago

-3(x – 14) + 9x = 6x + 4<br>

8090 [49]

Answer:

there are no solutions :3

no this is not a joke sjsjsj there are equations who has an answer and ones that are impossible to have a solution. use the no solution sign when you answer it or sum

Step-by-step explanation:

6 0

3 years ago

Pretty pls help me :3

lapo4ka [179]

Answer:

You can do 60 ÷ 6 = 10 groups.

Step-by-step explanation:

There are many ways to answer this and it's very simple.

6 and 10 is just one of the many factors of 60.

you can also do 60 ÷ 2 = 30 groups

3 0

3 years ago

Help please! Will pick the Brainliest.<br> (sorry if it’s sideways)

KatRina [158]

$(x^2y^8)^\frac{2}{3}=x^{2\cdot\frac{2}{3}}y^{8\cdot\frac{2}{3}}=x^{\frac{4}{3}}y^{\frac{16}{3}}=x^{1\frac{1}{3}}y^{5\frac{1}{3}}=x^1x^\frac{1}{3}y^5y^\frac{1}{3}=xy^5\sqrt[3]{x}\sqrt[3]{y}=\boxed{xy^5\sqrt[3]{xy}}$

Answer B)

$\frac{1}{3}\left(\frac{2}{15}x-\frac{2}{3}\right)>x+\frac{1}{5}\quad|\cdot3\\\\\\\frac{2}{15}x-\frac{2}{3}>3x+\frac{3}{5}\quad|\cdot15\\\\\\2x-\frac{30}{3}>45x+\frac{45}{5}\\\\2x-10>45x+9\\\\-10-9>45x-2x\\\\-19>43x\quad|:43\\\\\boxed{x$

Answer B)

$\sqrt{14xy}\cdot\sqrt{12}\cdot\sqrt{30y}=\sqrt{14\cdot12\cdot30}\cdot\sqrt{xy^2}=\sqrt{2\cdot7\cdot3\cdot4\cdot2\cdot3\cdot5}\cdot\sqrt{xy^2}=\\\\=\sqrt{2^2\cdot3^2\cdot2^2\cdot5\cdot7}\cdot\sqrt{xy^2}=2\cdot3\cdot2\cdot\sqrt{35}\cdot y\cdot\sqrt{x}=\boxed{12y\sqrt{35x}}$

3 0

4 years ago

Please that (b). I'm having some difficulties.

nata0808 [166]

Here's the answer.

Hope it helps u.

7 0

4 years ago